Question

Examples: Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) The graph of a function that is continuous, but not differentiable, at $x=2$.

(b) The graph of a function that is left and right differentiable, but not differentiable, at $x=3$.

(c) The graph of a function that is differentiable on the interval $[-1,1]$but not differentiable at the point $x=1.$

Short answer

(a). $$

\begin{aligned}

f^{\prime}(2) &=\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(-2)}{h} \\

&=-1

\end{aligned}

$$

(b). $\begin{aligned} f^{\prime}(3) &=\lim _{h \rightarrow 0^{+}} \frac{f(3+h)-f(3)}{h} \\ &=-1 \end{aligned}$

(c). $\begin{aligned} f^{\prime}(1) &=\lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h} \\ &=-1 \end{aligned}$

Step-by-step solution

Part (a) Step 1: Given information

Here the objective is to find the graph of a function that is continuous but not differentiable at $x=2$.

Part (a) Step 2: Simplification

Consider a function:

$$

f(x)= \begin{cases}x-2, & \text { if } x \geq 2 \\ -x+2, & \text { if } x<2\end{cases}

$$

The graph of the function will be:

From the graph it can be obacrved that the function is continuous at $x=2$ but it is not differentiable at $x=2$ as

$$

\begin{aligned}

f^{\prime}(2) &=\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(-2)}{h} \\

&=1

\end{aligned}

$$

But

$$

\begin{aligned}

f^{\prime}(2) &=\lim _{h \rightarrow 0^{+}} \frac{f(2+h)-f(-2)}{h} \\

&=-1

\end{aligned}

$$

Part (b) Step 1: Given information

Here the objective is to find the graph of a function that is left and right differentiable, but not differentiable at $x=3$.

Part (b) Step 2: Simplification

Consider a function:

$$

f(x)= \begin{cases}x-3, & \text { if } x \geq 3 \\ -x+3, & \text { if } x<3\end{cases}

$$

The graph of the function will be:

From the graph it can be observed that the function is differentiable at any point $c$ such that $c<3$ and $c>3$ but it is not differentiable at $x=3$ as

$$

\begin{aligned}

f^{\prime}(3) &=\lim _{h \rightarrow 0^{\circ}} \frac{f(3+h)-f(3)}{h} \\

&=1

\end{aligned}

$$

But

$$

\begin{aligned}

f^{\prime}(3) &=\lim _{h \rightarrow 0^{-}} \frac{f(3+h)-f(3)}{h} \\

&=-1

\end{aligned}

$$

Part (c) Step 1: Given information

Here the objective is to find the graph of a function that is differentiable on the interval $[-1,1]$ but not differentiable at the point $x=1$.

Part (c) Step 2: Simplification

Consider a function:

$$

f(x)= \begin{cases}x-1, & \text { if } x \geq 1 \\ -x+1, & \text { if } x<1\end{cases}

$$

The graph of the function will be:

From the graph it can be observed that the function is differentiable at any point $c$ such that $-1 \leq c<1$ but it is not differentiable at $x=1$ as

$$

\begin{aligned}

f^{\prime}(1) &=\lim _{h \rightarrow 0^{+}} \frac{f(1+h)-f(1)}{h} \\

&=1

\end{aligned}

$$

But

$$

\begin{aligned}

f^{\prime}(1) &=\lim _{h \rightarrow 0^{-}} \frac{f(1+h)-f(1)}{h} \\

&=-1

\end{aligned}

$$